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Math Help - inequalities

  1. #1
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    inequalities

    Please check my work.thank you so much
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  2. #2
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    Quadratic Inequalities are always easiest to solve if you complete the square.

    \displaystyle x^2 - 2x - 3 \leq 0

    \displaystyle x^2 - 2x + (-1)^2 - (-1)^2 - 3 \leq 0

    \displaystyle (x - 1)^2 - 4 \leq 0

    \displaystyle (x - 1)^2 \leq 4

    \displaystyle \sqrt{(x-1)^2} \leq \sqrt{4}

    \displaystyle |x-1| \leq 2

    \displaystyle -2 \leq x - 1 \leq 2

    \displaystyle -1 \leq x \leq 3.
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  3. #3
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    Another good way to solve quadratic inequalities (and other complicated inequalities) is to solve the associated equation.

    For example, to solve \displaystyle x^2 - 2x - 3 \leq 0, first solve \displaystyle x^2 - 2x - 3= 0, by factoring, (x- 3)(x+ 1)= 0 so x= 3 and x= -1. The point is this: a polynomial is a "continuous" function which means that it in order to go from positive to negative, it must pass through 0 and that can happen only at x= -1 and x= 3 which divide the real line into three intervals.

    To decide which interval is ">0" and which "< 0" you can use either of two methods:
    1) Choose a number in each interval and check the sign. x= -2 is less than -1 and (-2)^2- 2(-2)- 3= 4+ 4- 3= 5> 0 so the polynomial is greater than 0 for all x< -2. x= 0 is between -1 and 3 and (0)^2- 2(0)- 3=-3< 0 so the polynomial is less than 0 for all x between -1 and 3. Finally, x= 4 is larger than 3 and (4)^2- 2(4)- 3= 5> 0 so the polynomial is greater than 0 for all x greater than 3. Since the inequality say " \le 0", the solution set is -1\le x\le 3.

    2) Notice that x+ 1 is negative for x< -1 and positive for x> -1 and that x- 3 is negative for x< 3 and positive for x> 3. If x< -1, it is also less than 3 so both factors are negative and their product is positive. If -1< x< 3, one factor, x+1, is positive but x- 3 is still negative. The product of one positive and one negative factor is negative. Finally, if x> 3, both factors are positive and their product is positive. That gives the same solution set as before and (fortunately!) the same solution as Prove It.
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  4. #4
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    Quote Originally Posted by jam2011 View Post
    Please check my work.thank you so much
    x^2-2x-3\ \le\ 0

    (x-3)(x+1)\le\ 0

    Your Case 1:

    If x-3 \ge\ 0, then x+1=(x-3)+4 must be \ge\ 0 since it is 4 greater than (x-3)

    No luck there...

    Case 2:

    If x-3\ \le\ 0, then x\ \le\ 3 and x+1\ \ge\ 0 only for x\ \ge\ -1


    Hence -1\ \le\ x\ \le\ 3

    so your answer is correct and the diagrams help illustrate the intersections.


    Yet another way to proceed is....

    x^2-2x\ \le\ 3\Rightarrow\ x(x-2)\ \le\ 3

    The factors of 3 that differ by 2 are 1 and 3 or -1 and -3

    x(x-2)=3 for x=3 and for x=-1

    If x<-1, then x(x-2)>3

    If x>3, then x(x-2)>3

    The solution is -1\ \le\ x\ \le\ 3
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  5. #5
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    x^2 - 2x - 3 \le 0

    (x + 1)(x - 3) \le 0

    think about the graph of the quadratic ... a parabola that opens upward with two zeros, x = -1 and x = 3

    the section of the graph below the x-axis (the less than or equal to section) lies between and includes the two zeros.
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  6. #6
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    nice job !!! thank you so much...i learn a lot.there are many ways to find and solve this problem. i owe you a lot...
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